[greenmoon]

It's 1.02 liters of KOH and 0.97 liters of HNO3.

[greenmoon]

Did I calculate it right?

[sjb]

It's 1.02 liters of KOH and 0.97 liters of HNO3.

So now you have an excess of 50 ml, 0.35M KOH. How many moles is this?

[greenmoon]

It's 0.89 moles.

[sjb]

It's 0.89 moles.

Is 0.89 mol in 2 litres of solution the right concentration to give the pH you require?

[greenmoon]

No....

[sjb]

No....

OK, so we've made an error somewhere.

If you have V_B litres of KOH, how many litres of acid do we have, to make a total solution of 2 litres?

If we have V_B litres of 0.35 M KOH, how many moles of KOH do we have?

From the first part, how many moles of 0.35 M HNO₃ do we have?

What should the difference of these last two be, to give a pH of 12 at the end?

[greenmoon]

V_bliters? I don't get it...you mean like if V_b is the unknown?

[sjb]

V_bliters? I don't get it...you mean like if V_b is the unknown?

Yes, you have two unknowns, V_B litres of base, and V_A of acid.

You can formulate several equations using these to eventually get the data you want.

For instance, as the total volume is 2 litres, V_A + V_B = 2.

[greenmoon]

Ah,got it now.
Well if it's V_a+V_b=2,then V_a=2-V_b.
If we have 0.35 KOH and 0.35 HNO3 V_b=x/0.35 and V_a=x/0.35 but if pH=12 and the volume is 2 L then the difference between them in 0.02.So V_b=x+0.02/0.35 and V_a=x/0.35.So if I add those two I get the same equation I got before....

[sjb]

Maybe it's just a rounding issue then. Can you post the rest of your working, but don't round figures off prematurely.

[greenmoon]

x/0.35+x+0.02/0.35=2
(2x+0.02)/0.35=2
2x=0.68
x=0.34
So the volume of HNO3=0.34 and the volume of KOH=0.36.And I've done no rounding.
The 0.35 of KOH is though a bit hard to read cause it's a copy it might be 0.25.I'll try and re analyze.
Many thanks.

[sjb]

x/0.35+x+0.02/0.35=2
(2x+0.02)/0.35=2
2x=0.68
x=0.34
So the volume of HNO3=0.34 and the volume of KOH=0.36.And I've done no rounding.
The 0.35 of KOH is though a bit hard to read cause it's a copy it might be 0.25.I'll try and re analyze.
Many thanks.

No, I think the number of moles of KOH and HNO₃ are as you've calculated, check your units.